Probability of Ugrading Your Poker Hand

If you are playing a game of Five Card-Draw with a friend out of a standard deck 52 card deck and you have the following hand (neglecting suits for this example) 3,3,3,5,7. You decide to discard the "5" and draw from the deck. What is the probability that you upgrade to Four of a Kind ( 3,3,3,3,7)?

I'm having a bit of trouble wrapping my head around the following... Does the fact that your opponent has 5 of the 47 remaining out of the draw deck ( leaving 42 in the deck from which you can choose) effect the probability of you drawing the remaining "3" or not?

Originally posted by @joe-shmo
If you are playing a game of Five Card-Draw with a friend out of a standard deck 52 card deck and you have the following hand (neglecting suits for this example) 3,3,3,5,7. You decide to discard the "5" and draw from the deck. What is the probability that you upgrade to Four of a Kind ( 3,3,3,3,7)?

I'm having a bit of trouble wrapping my head around th ...[text shortened]... deck from which you can choose) effect the probability of you drawing the remaining "3" or not?

Thing is, your opponent could hold the remaining 3 which makes your chances 0/42.
His chances of having the remaining 3 is 5/42.
If he doesn't have the 3 your chances are 1/42.
I'm sure there's a way to combine this information to come up with the answer but I don't know how to do it.
I'd be interested to see the answer.

Originally posted by @venda
Thing is, your opponent could hold the remaining 3 which makes your chances 0/42.
His chances of having the remaining 3 is 5/42.
If he doesn't have the 3 your chances are 1/42.
I'm sure there's a way to combine this information to come up with the answer but I don't know how to do it.
I'd be interested to see the answer.

Surely, from where you are (knowing 5 cards), the chance of him having the remaining 3 is 5/47?

Anyway, it is unimportant to your odds calculation. There are 47 cards not in your hand, that you don't know, and you want just 1, so prob of it being the top card is 1/47 to you.

Think of it that you are concentrating on the top card of the pack. Just one card. If the 3 is anywhere else (opponents hand, bottom of pack, middle) , you lose.

Chance of it being top card to you is 1/47

Originally posted by @blood-on-the-tracks
Surely, from where you are (knowing 5 cards), the chance of him having the remaining 3 is 5/47?

Anyway, it is unimportant to your odds calculation. There are 47 cards not in your hand, that you don't know, and you want just 1, so prob of it being the top card is 1/47 to you.

Think of it that you are concentrating on the top card of the pac ...[text shortened]... pponents hand, bottom of pack, middle) , you lose.

Chance of it being top card to you is 1/47

I agree with you. The probability is 1/47.

However, I would like to pin down my point of confusion if possible.

Lets start with you being dealt the five cards (3,3,3,5,7) and the deck. Player 2 has no cards at this time.

The probability that player 2 has the remaining "3" = 0
The probability the deck has the "3" = 1

Now, player 2 draws five cards without looking the probability that he has the now has the "3" = 5/47
The deck holds the "3" with probability = 1 - 5/47 = 42/47

Now, allow player 2 to continue drawing cards, until he draws the last card. The probability that he has the "3" after this = 1
The probability the deck has the "3" = 0 (it doesn't exist anymore).

So, as you allow player 2 to draw cards the probability he obtains the "3" increases from 0 to 1 while the probability the deck has the "3" goes from 1 to 0.

How is it that as probability the "3" remains in the deck ( the very thing you are picking out of ) decreases to 0, but the probability of you choosing the "3" at every draw up until the very last card is drawn out of that deck remains constant at 1/47?

So...what is the conflation here, or is it the whole line of reasoning above that is incorrect?

Originally posted by @joe-shmo
I agree with you. The probability is 1/47.

However, I would like to pin down my point of confusion if possible.

Lets start with you being dealt the five cards (3,3,3,5,7) and the deck. Player 2 has no cards at this time.

The probability that player 2 has the remaining "3" = 0
The probability the deck has the "3" = 1

Now, player 2 draws five ca ...[text shortened]...
So...what is the conflation here, or is it the whole line of reasoning above that is incorrect?

It doesn't matter if the 3 is in the deck or in the opponent's hand. All that matters is if it is the top card when it's your turn to draw. The deck's probability will only go to 0, and the opponent's to 1, when the last card is drawn. In that case you cannot draw a card so there is no 1/47.

Of course it's out of 47 not 42.I thought I had put 47.
I'm not familiar with the game but yes , if it's just about the top card in the deck 1/47 is correct.

Originally posted by @joe-shmo
...
How is it that as probability the "3" remains in the deck ( the very thing you are picking out of ) decreases to 0, but the probability of you choosing the "3" at every draw up until the very last card is drawn out of that deck remains constant at 1/47?...

But the probability of you drawing that '3' from the deck keeps increasing. If there were only one card left in the deck and you still haven't found that '3', the probability has increased to 1.

Originally posted by @apathist
But the probability of you drawing that '3' from the deck keeps increasing. If there were only one card left in the deck and you still haven't found that '3', the probability has increased to 1.

Deck or 'opponents hand' , doesn't affect the odds.

Unless your opponent reveals what card he has after each draw, which , of course, doesn't happen

You need 1 card, there are 47 unknown, so odds = 1/47

(Don't discount the possibility of getting a v strong 'full house'!!)

Originally posted by @blood-on-the-tracks
Deck or 'opponents hand' , doesn't affect the odds.

Unless your opponent reveals what card he has after each draw, which , of course, doesn't happen

You need 1 card, there are 47 unknown, so odds = 1/47

(Don't discount the possibility of getting a v strong 'full house'!!)

If you are playing against my flush then your chances of a full-house increase greatly.
This has been demonstrated experimentally many times. (The "Wolfgang Effect" )
https://www.ideapod.com/idea/Interesting-math-oddities/2468069

Originally posted by @wolfgang59
If you are playing against my flush then your chances of a full-house increase greatly.
...

Of course what cards are held by others affects what cards are available. Are you sharing your hole cards with others?

Originally posted by @apathist
Of course what cards are held by others affects what cards are available. Are you sharing your hole cards with others?

Just messing.
When I get a good hand there is always one better out there!

Originally posted by @blood-on-the-tracks
Deck or 'opponents hand' , doesn't affect the odds.

Unless your opponent reveals what card he has after each draw, which , of course, doesn't happen

You need 1 card, there are 47 unknown, so odds = 1/47

(Don't discount the possibility of getting a v strong 'full house'!!)

you are correct

Originally posted by @venda
Thing is, your opponent could hold the remaining 3 which makes your chances 0/42.

The next card in the deck could be the one you want, making your chances 100%

Originally posted by @apathist
The next card in the deck could be the one you want, making your chances 100%

Only if you exchange the word "could" in your statement For "is"

Originally posted by @venda
Only if you exchange the word "could" in your statement For "is"

That works for a shuffled deck, but not for an unrolled die.